 Hello and welcome to the session. The given question says, form the differential equation of the family of curves y is equal to a sin x plus b where a and b are arbitrary constants. Let's start with the solution and we are given y is equal to a sin x plus b where a and b are arbitrary constants. Now since as we can see here we have two arbitrary constants a and b therefore we will differentiate this equation two times and first let us differentiating it once. Let us denote this by equation number one. So differentiating equation number one with respect to x we get on the left hand side divided by dx is equal to a into derivative of sin x plus b which is cos into x plus b and derivative of x plus b is one plus sin plus b into derivative of a and a is a constant so it is equal to zero. Therefore we have divided by dx is equal to a cos x plus b. Let this be equation number two. Now as we have already discussed there are two constants a and b so we will differentiate it twice. So again differentiating it with respect to x and now we will differentiate equation number two with respect to x. On differentiating the left hand side we have d square y divided by dx square and on the right hand side we have a and derivative of cos x plus b is minus sin x plus b into derivative of x plus b is one and we have plus cos x plus b and derivative of a is zero. So this is further equal to d square y divided by dx square is equal to minus a sin x plus b plus b sy. So this implies d square y divided by dx square is equal to minus y and this is from equation number one or this further implies that d square y divided by dx square plus y is equal to zero which is the required equation. So this completes the equation by and take care.